Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $q = \dfrac{3y^3 + 3y^2 - 6y}{3y^3 - 9y^2 - 30y} \times \dfrac{y + 4}{-y + 1} $
Solution: First factor out any common factors. $q = \dfrac{3y(y^2 + y - 2)}{3y(y^2 - 3y - 10)} \times \dfrac{y + 4}{-(y - 1)} $ Then factor the quadratic expressions. $q = \dfrac {3y(y + 2)(y - 1)} {3y(y + 2)(y - 5)} \times \dfrac {y + 4} {-(y - 1)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac { 3y(y + 2)(y - 1) \times (y + 4)} { 3y(y + 2)(y - 5) \times -(y - 1)} $ $q = \dfrac {3y(y + 2)(y - 1)(y + 4)} {-3y(y + 2)(y - 5)(y - 1)} $ Notice that $(y + 2)$ and $(y - 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {3y\cancel{(y + 2)}(y - 1)(y + 4)} {-3y\cancel{(y + 2)}(y - 5)(y - 1)} $ We are dividing by $y + 2$ , so $y + 2 \neq 0$ Therefore, $y \neq -2$ $q = \dfrac {3y\cancel{(y + 2)}\cancel{(y - 1)}(y + 4)} {-3y\cancel{(y + 2)}(y - 5)\cancel{(y - 1)}} $ We are dividing by $y - 1$ , so $y - 1 \neq 0$ Therefore, $y \neq 1$ $q = \dfrac {3y(y + 4)} {-3y(y - 5)} $ $ q = \dfrac{-(y + 4)}{y - 5}; y \neq -2; y \neq 1 $